Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Test - Page 293: 10


$\theta=90^o\text{ or } \theta = 270^o$

Work Step by Step

Subtract $\sin^2{\theta}$ to both sides: $0=\cos^2{\theta} +1 - \sin^2{\theta} \\0=\cos^2{\theta}-\sin^2{\theta}+1$ RECALL: $\cos{(2\theta)}= \cos^2{\theta}-\sin^2{\theta}$ Use the identity above to obtain: $0=(\cos^2{\theta}-\sin^2{\theta})+1 \\0=\cos{(2\theta)}+1 \\-1=\cos{(2\theta)}$ Note that $\cos{180^o}=-1$ and $\cos{540^o}=-1$ Thus, $2\theta=180^o \text{ or } 2\theta=540^o \\\theta=\frac{180^o}{2} \text{ or } \theta = \frac{540^o}{2} \\\theta=90^o\text{ or } \theta = 270^o$
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